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(B) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$. Freezi

Vedclass · Accepted Answer

$a-q, b-p, c-s, d-r$

Vedclass · Answer

(B) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$. Freezing point $T_f = T_f^0 - \Delta T_f = 273 - (i 	imes K_f 	imes 0.1)$. For sucrose,$i=1$,$\Delta T_f = 273 - 272 = 1 \, K$. Thus,$K_f 	imes 0.1 = 1$. $T_f = 273 - i$. For $a. \ BaCl_2$ $(i=3)$: $T_f = 273 - 3 = 270 \, K$ $(q)$. For $b. \ NaCl$ $(i=2)$: $T_f = 273 - 2 = 271 \, K$ $(p)$. For $c. \ K_3[Fe(CN)_6]$ $(i=4)$: $T_f = 273 - 4 = 269 \, K$ $(s)$. For $d. \ Al_2(SO_4)_3$ $(i=5)$: $T_f = 273 - 5 = 268 \, K$ $(r)$. Therefore,the correct match is $a-q, b-p, c-s, d-r$.

Column-$I$ (Various solutions)	Column-$II$ (Freezing point)
$a$. $0.1 \, M \ BaCl_2$ solution	$p$. $271 \, K$
$b$. $0.1 \, M \ NaCl$ solution	$q$. $270 \, K$
$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution	$r$. $268 \, K$
$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution	$s$. $269 \, K$

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